InterviewSolution
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InterviewSolution
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A coil of inductance `0.1 H` is connected to `50 V, 100Hz` generator and current is found to be `0.5A`. The potential difference across resistance of coil is:A. `15 V`B. `20 V`C. `25 V`D. `39 V` |
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Answer» Correct Answer - 4 `I = (E)/(Z) , 0.5 = (50)/(Z) = 100 Omega` `Z^(2) = R^(2) + omega^(2) L^(2)`, then `R = 78 Omega` Now `V_(R ) = sqrt(V_(LR)^(2) = V_(L)^(2)) = 39 V , [V_(R)^(2) + V_(L)^(2) = V_(LR)^(2)]` |
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